Mathematical Reflections 2012, Issue 5 - Problem U243

Problem:
Let $f:(a,b) \longrightarrow \mathbb{R}$ be a differentiable function such that $f'(a)=f'(b)=0$ and with the
property that there is a real valued function $g$ for which $g(f'(x))=f(x)$ for all $x$ in $\mathbb{R}$.
Prove that $f$ is constant.

Proposed by Mihai Piticari and Sorin Radulescu.

Solution:
Since $f$ is differentiable in $(a,b)$ and $f'(a)=f'(b)=0$, $f$ is continuous in $[a,b]$. By Weierstrass's Theorem there exist maximum $M$ and minimum $m$ in $[a,b]$. If these are both attained at $a$ and $b$, $f$ is constant. If one of these is not attained at $a$ or $b$, then there exists $c \in (a,b)$ such that $f(c)=M$ or $f(c)=m$. Since $f$ is differentiable $f'(c)=0$, so $g(0)=f(c)=f(a)=f(b)$, which means that $f$ is constant in $(a,b)$.

Mathematical Reflections 2012, Issue 5 - Problem U241

Problem:
Let $a>b$ be positive real numbers. Prove that $$c_n=\dfrac{\sqrt[n+1]{a^{n+1}-b^{n+1}}}{\sqrt[n]{a^n-b^n}}$$
is a decreasing sequence and find its limit.

Proposed by Ivan Borsenco.

Solution:
Let $c_n=\dfrac{\sqrt[n+1]{1-x^{n+1}}}{\sqrt[n]{1-x^n}}$, where $x=b/a < 1$. We first prove that $c_n > 1$. Let $b_n=c^{n(n+1)}_n$. Then $$b_n(x)=\dfrac{(1-x^{n+1})^n}{(1-x^n)^{n+1}}$$ and $$b'_n(x)=\dfrac{n(n+1)x^{n-1}(1-x^{n+1})^{n-1}(1-x^n)^n(1-x)}{(1-x^n)^{2(n+1)}}.$$ So, $b'_n(x) > 0$ for all $x \in (0,1)$, so $b_n(x)$ is increasing in $(0,1)$ and $b_n(x) > b_n(0)=1$, i.e. $c^{n(n+1)}_n > 1$, which yields $c_n > 1$. Now, we prove that $c_n$ is a decreasing sequence. We have $$\dfrac{b_{n-1}}{b_n}=\left(\dfrac{(1-x^n)^2}{(1-x^{n-1})(1-x^{n+1})}\right)^n > 1,$$ so $c_{n-1}^{(n-1)n}>c_n^{n(n+1)}=c_n^{(n-1)n}c_n^{2n}>c_n^{(n-1)n}$, i.e. $c_{n-1}>c_n$. Moreover,
$$\lim_{n \to \infty} \log c_n=\lim_{n \to \infty} \left[\dfrac{1}{n+1}\log (1-x^{n+1})-\dfrac{1}{n}\log(1-x^n)\right]=0,$$ so $$\lim_{n \to \infty} c_n= \lim_{n \to \infty} e^{\log c_n}=1.$$

Mathematical Reflections 2012, Issue 5 - Problem S243

Problem:
A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.

Proposed by Iurie Boreico.

Solution:
Let $m$ be the number of boys and $n$ be the number of girls. There are $\displaystyle{m \choose 2}$ possibile pairs of boys and $\displaystyle{n \choose 2}$ possible pairs of girls. Suppose that $\displaystyle{m \choose 2} > \displaystyle{n \choose 2}$. Then, there are two distinct pairs of boys which dance with a pair of girls, but this means that there are at least three boys which dance with two girls, contradiction. With the same reasoning we can prove that it can't be $\displaystyle{m \choose 2} < \displaystyle{n \choose 2}$, so $\displaystyle{m \choose 2} = \displaystyle{n \choose 2}$, i.e. $m=n$.

Mathematical Reflections 2012, Issue 5 - Problem S241

Problem:
Let $p$ and $q$ be odd primes such that $\dfrac{p^3-q^3}{3} \geq 2pq+3.$ Prove that
$$\dfrac{p^3-q^3}{4} \geq 3pq+16.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $p>q$, so $p-2 \geq q$. It must be $p-2 \neq q$, otherwise we would have $$p^3-q^3=p^3-(p-2)^3=6p^2-12p+8=6pq+8,$$ which contradicts the given inequality. So $p-4 \geq q$ and $$p^3-q^3 \geq p^3-(p-4)^3=12p^2-48p+64=12p(p-4)+64 \geq 12pq+64,$$ which gives the conclusion.

Mathematical Reflections 2012, Issue 5 - Problem J245

Problem:
Find all triples $(x,y,z)$ of positive real numbers satisfying simultaneously the inequalities $x+y+z-2xyz \leq 1$ and $$xy+yz+zx+\dfrac{1}{xyz} \leq 4.$$

Proposed by Titu Andreescu.

Solution:
Using HM-AM Inequality, from the first inequality we have
$$\dfrac{9xyz}{xy+yz+zx} \leq x+y+z \leq 1+2xyz,$$ which gives $\dfrac{9xyz}{2xyz+1} \leq xy+yz+zx$. From the second inequality, we have $$\dfrac{9xyz}{2xyz+1}+\dfrac{1}{xyz} \leq 4.$$ Putting $t=xyz>0$ and clearing the denominators, we obtain $$9t^2+2t+1 \leq 4t(2t+1),$$ i.e. $(t-1)^2 \leq 0$, which gives $t=1$. So, the first inequality is $x+y+z \leq 3xyz$, but by AM-GM Inequality $3xyz \leq x+y+z$, which implies $x+y+z=3xyz$. The equality holds if and only if $x=y=z=1$, so the only triple which satisfies the two inequalities is $(1,1,1)$.

Mathematical Reflections 2012, Issue 5 - Problem J244

Problem:
Let $a$ and $b$ positive real numbers. Prove that
$$1 \leq \dfrac{\sqrt[n]{a^n+b^n}}{\sqrt[n+1]{a^{n+1}+b^{n+1}}} \leq \sqrt[n(n+1)]{2}.$$

Proposed by Ivan Borsenco.

Solution:

Since we can collect $a$ or $b$ in the given expression, it is sufficient to prove that
$$1 \leq \dfrac{\sqrt[n]{1+x^n}}{\sqrt[n+1]{1+x^{n+1}}} \leq \sqrt[n(n+1)]{2} \qquad \forall x \in \mathbb{R}^+,$$ or equivalently
$$1 \leq \dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n} \leq 2, \qquad \forall x \in \mathbb{R}^+.$$ Put $f(x)=\dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n}$. Then, $$f'(x)=\dfrac{n(n+1)x^{n-1}(1+x)^n(1+x^{n+1})^{n-1}(1-x)}{[(1+x^{n+1})^n]^2},$$ and $f'(x) > 0$ if and only if $x<1$, i.e. $f(1)$ is a maximum. Moreover, $f(0)=1$ and $\lim_{x \to +\infty} f(x)=1^+$, so $f(0) \leq f(x) \leq f(1)$ for all positive real numbers $x$, which gives $1 \leq f(x) \leq 2$ for all positive real numbers $x$.

Mathematical Reflections 2012, Issue 5 - Problem J243

Problem:
Let $a,b,c$ be real numbers such that $$\left(-\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{c}{6}\right)^3+\left(\dfrac{a}{3}+\dfrac{b}{6}-\dfrac{c}{2}\right)^3+\left(\dfrac{a}{6}-\dfrac{b}{2}+\dfrac{c}{3}\right)^3=\dfrac{1}{8}.$$ Prove that $$(a-3b+2c)(2a+b-3c)(-3a+2b+c)=9.$$

Proposed by Titu Andreescu.

Solution:
From the given equality, we have $$(-3a+2b+c)^3+(2a+b-3c)^3+(a-3b+2c)^3=27.$$
Let $x=-3a+2b+c, y=2a+b-3c, z=a-3b+2c$. Since $x+y+z=0$, using the well known identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$ we get $x^3+y^3+z^3=3xyz$, i.e.
$$3(-3a+2b+c)(2a+b-3c)(a-3b+2c)=27,$$ which gives $(-3a+2b+c)(2a+b-3c)(a-3b+2c)=9$.

Mathematical Reflections 2012, Issue 5 - Problem J242

Problem:
Let $ABC$ be a triangle and let $D,E,F$ be the feet of the altitudes from $A,B,C$ to the
sides $BC,CA,AB$, respectively. Let $X,Y,Z$ be the midpoints of segments $EF,FD,DE$ and let $x,y,z$ be the perpendiculars from $X,Y,Z$ to $BC,CA$, and $AB$, respectively. Prove that the lines $x,y,z$ are concurrent.

Proposed by Cosmin Pohoata.

Solution:

Obviously, the lines $x,y,z$ are parallel to the altitudes $AD,BE,CF$ respectively. Since the orthocenter of the triangle $ABC$ is the incenter of the orthic triangle $DEF$, the lines $AD,BE,CF$ are bisectrices of the angle $D,E,F$ of the triangle $DEF$ respectively. Moreover, since $X,Y,Z$ are midpoints of the segments $EF,FD,DE$, then $XY,YZ,ZX$ are parallel to $DE,EF,FD$. The angle between the line $XY$ and the line $x$ and the angle $ADE$ are equal (alternate interior angles) and so are the angle between the line $XZ$ and the line $x$ and the angle $ADF$ (alternate exterior angles), so the line $x$ is the bisectrix of the angle $X$ of the triangle $XYZ$. Likewise, the lines $y$ and $z$ are bisectrices of the angles $Y,Z$ respectively. So the lines $x,y,z$ are concurrent since their point of intersection is the incenter of the triangle $XYZ$. 

Mathematical Reflections 2012, Issue 5 - Problem J241

Problem:
Determine all positive integers that can be represented as $$\dfrac{ab+bc+ca}{a+b+c+\min (a,b,c)}$$ for some positive integers $a,b,c$.

Proposed by Titu Andreescu.

Solution:

We claim that every positive integer $n$ can be represented in this form. Suppose $a \geq b \geq c$. Let $c=1, b=n$. Then
$$\dfrac{ab+bc+ca}{a+b+c+\min (a,b,c)}=\dfrac{a(n+1)+n}{a+n+2}=n+1-\dfrac{n^2+2n+2}{a+n+2},$$ so it suffices to choose $a=n^2+n$. 

Mathematical Reflections 2012, Issue 4 - Problem O235

Problem:
Solve in integers the equation $$xy-7\sqrt{x^2+y^2}=1.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $xy \geq 1$. By symmetry, we reduce to $x,y > 0$. Rewriting the equation in the form $(xy-1)^2=49(x^2+y^2)$, we put $xy=7t+1, t \geq 0$, so that $x^2+y^2=t^2$. Then $$(x+y)^2=x^2+y^2+2xy=t^2+2(7t+1)=(t+7)^2-47,$$ which gives $$(t+7-x-y)(t+7+x+y)=47.$$
So, we must solve the systems
$$\left\{\begin{array}{rcl} t+7-x-y & = & 1 \\ t+7+x+y & = & 47 \end{array} \right. \qquad \left\{\begin{array}{rcl} t+7-x-y & = & -47 \\ t+7+x+y & = & -1 \end{array} \right.$$
Solving the first, we get $t=17$ and $x+y=23$, solving the second we get $t=-31<0$, i.e. no solution. So we have $x+y=23, xy=120$, which gives $x=15,y=8$ and $x=8, y=15$. In conclusion, the solutions are $$(15,8),(8,15),(-15,-8),(-8,-15).$$

Mathematical Reflections 2012, Issue 4 - Problem U237

Problem:
Let $\mathcal{H}$ be a hyperbola with foci $A$ and $B$ and center $O$. Let $P$ be an arbitrary point
on $\mathcal{H}$ and let the tangent of $\mathcal{H}$ through $P$ cut its asymptotes at $M$ and $N$. Prove that $PA + PB = OM + ON$.

Proposed by Luis Gonzalez.

Solution:

Let $\mathcal{H}: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, so that $O=(0,0)$. By symmetry, suppose that $P$ is on the first quadrant. Then $P=\left(k,\dfrac{b}{a}\sqrt{k^2-a^2}\right)$, where $k \geq a$ is an arbitrary real number. The equation of the tangent of $\mathcal{H}$ at $P$ is $\dfrac{kx}{a^2}-\dfrac{\dfrac{b}{a}\sqrt{k^2-a^2}y}{b^2}=1$, so intersecting this line with the asymptotes $y=\pm \dfrac{b}{a}x$ we get the points $$M=\left(k+\sqrt{k^2-a^2},\dfrac{b}{a}(k+\sqrt{k^2-a^2})\right), \qquad N=\left(k-\sqrt{k^2-a^2},-\dfrac{b}{a}(k-\sqrt{k^2-a^2})\right).$$ Then
$$OM+ON=\dfrac{2k}{a}\sqrt{a^2+b^2}=2ke,$$ where $e=\sqrt{a^2+b^2}/a$ is the eccentricity of the hyperbola. Since the foci have coordinates $A=(-c,0), B=(c,0)$ where $c=\sqrt{a^2+b^2}$, we have
$$\begin{array}{lcl}PA+PB & = &\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}+\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)} \\ & = & \dfrac{4kc}{\sqrt{(k+c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}-\sqrt{(k-c)^2+\dfrac{b^2}{a^2}(k^2-a^2)}}\\ &=&\dfrac{2kc}{a}\\&=&2ke \end{array}$$
and the statement follows.

Mathematical Reflections 2012, Issue 4 - Problem U236

Problem:
Let $f(X)$ be an irreducible polynomial in $\mathbb{Z}[X]$. Prove that $f(XY)$ is irreducible in $\mathbb{Z}[X,Y]$.

Proposed by Mircea Becheanu.

Solution:

If $f(X)$ is a constant polynomial, the statement is trivial. So, assume that $\deg f(X) > 0$. Since $f(X)$ is irreducible, then must be primitive. Assume that $f(XY)$ is reducible in $\mathbb{Z}[X,Y]$. Then, there are two non constant polynomials $g(X,Y), h(X,Y) \in \mathbb{Z}[X,Y]=\mathbb{Z}[X][Y]$ such that $f(XY)=g(X,Y)h(X,Y)$. It cannot be $f(XY)=g(X)h(Y)$, otherwise $f(0)=g(0)h(Y)$, so $h(Y)$ must be constant. Then both $g(X,Y)$ and $h(X,Y)$ must contain both $X$ and $Y$. But $$f(X)=g(X,1)h(X,1)=\overline{g}(X)\overline{h}(X), \qquad \overline{g}(X), \overline{h}(X) \in \mathbb{Z}[X]$$ is the product of two non constant polynomials, i.e. $f(X)$ is reducible, contradiction.

Mathematical Reflections 2012, Issue 4 - Problem S235

Problem:
Solve the equation $$\dfrac{8}{\{x\}}=\dfrac{9}{x}+\dfrac{10}{[x]},$$ where $[x]$ and $\{x\}$ denote the greatest integer less or equal than $x$ and the fractional part of $x$, respectively.

Proposed by Titu Andreescu.

Solution:

Clearly $x \notin \mathbb{Z}, x \not \in (0,1)$ and from $\dfrac{8}{\{x\}} > 0$, we gather $x > 1$. Since $0 < \{x\} < 1$ and $0 < [x] \leq x$, we have $$8 < \dfrac{8}{\{x\}} \leq \dfrac{19}{[x]},$$ so $[x]=1,2$. Clearing the denominators and using the fact that $\{x\}=x-[x]$ we obtain the equation $$10x^2-9x[x]-9[x]^2=0.$$
(i) If $[x]=1$, we get $10x^2-9x-9=0$, which gives $x=\dfrac{3}{2},-\dfrac{3}{5}$, i.e. $x=\dfrac{3}{2}$. 

(ii) If $[x]=2$, we get $10x^2-18x-36=0$, which gives $x=3,-\dfrac{6}{5}$, i.e. no solution.

So, $x=\dfrac{3}{2}$ is the only solution to the given equation.

Mathematical Reflections 2012, Issue 4 - Problem J239

Problem:
Let $a$ and $b$ be real numbers so that $2a^2 + 3ab + 2b^2 \leq 7$. Prove that $$\max\{2a+b,2b+a\} \leq 4.$$

Proposed by Titu Andreescu.

Solution:

Suppose that $a \geq b$, so that $\max\{2a+b,2b+a\}=2a+b$. We argue by contradiction. Suppose that for some real numbers $a,b$ we have $2a+b>4$, i.e. $a>\dfrac{4-b}{2}$. Then
$$7 \geq 2a^2 + 3ab + 2b^2 > b^2+2b+8,$$ which gives $(b+1)^2 < 0$, contradiction.

Mathematical Reflections 2012, Issue 4 - Problem J237

Problem:
Prove that the diameter of the incircle of a triangle $ABC$ is equal to $\dfrac{AB-BC+CA}{\sqrt{3}}$ if and only if $\angle BAC = 60^{\circ}$.

Proposed by Titu Andreescu.

Solution:
Let $r$ be the radius of the incircle of $\triangle ABC$ and let $\alpha=\angle BAC$. Clearly $0^{\circ} < \alpha < 180^{\circ}$ and $$2r=\dfrac{2AB\cdot CA \sin \alpha}{AB+BC+CA}.$$ Then $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff AB^2+CA^2+2AB\cdot CA - BC^2=2\sqrt{3}AB \cdot CA \sin \alpha.$$ Since $BC^2=AB^2+CA^2-2AB\cdot CA \cos \alpha$, we obtain $$2r=\dfrac{AB-BC+CA}{\sqrt{3}} \iff 1 + \cos \alpha = \sqrt{3} \sin \alpha,$$ i.e. if and only if $\sin (\alpha-30^{\circ})=\dfrac{1}{2},$ which gives $\alpha=60^{\circ}$.

Mathematical Reflections 2012, Issue 4 - Problem J235

Problem:
In the equality $\sqrt{ABCDEF}=DEF$, different letters represent different digits. Find the six-digit number $ABCDEF$.

Proposed by Titu Andreescu.

Solution:
From the given equality, it must be $ABCDEF^2-DEF \equiv 0 \pmod {1000}$, and since $ABCDEF \equiv DEF \pmod{1000}$, we have $DEF^2-DEF=DEF(DEF-1) \equiv 0 \pmod{1000}$. Since $DEF$ and $DEF-1$ are coprime and $1000=2^3\cdot5^3$, one between this two numbers must be odd and divisibile by $5^3$ and the other must be divisible by $2^3$. Since $DEF$ and $DEF-1$ are three digit numbers, $DEF \in \{125,375,625,875\}$ or $DEF-1 \in \{125,375,625,875\}$. In the first case, $DEF-1$ is divisible by $8$ if and only if $DEF=625$, in the second case $DEF$ is divisible by $8$ if and only if $DEF-1=375$. So, $DEF=625,376$, but
$$625^2=390625 \qquad 376^2=141376,$$ i.e. the only number which satisfies the required conditions is $625$.